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For $\mathbb{F}$ a finite abelian extension of $\mathbb{Q}$, recall that the conductor $c(\mathbb{F})$ of $\mathbb{F}$ is the smallest positive integer such that $\mathbb{F}$ is contained in the $c(\mathbb{F})$-th cyclotomic field. Define $\ell(\mathbb{F})=\log_p(c(\mathbb{F}))_p,$ where $n_p$ denotes the highest $p$-power dividing $n$. My question is kind of vague:

What can we say about the relation between $\mathbb{F}_1$ and $\mathbb{F}_2$ when we know $\ell(\mathbb{F}_1)=\ell(\mathbb{F}_2)$?

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  • $\begingroup$ Is $p$ fixed? Regardless of how this is interpreted, I think the answer is "not much". $\endgroup$ Commented Oct 8, 2022 at 22:51
  • $\begingroup$ Yes, $p$ is fixed. $\endgroup$ Commented Oct 9, 2022 at 0:49
  • $\begingroup$ What kind of thing would you hope to say? For instance we have $\ell(\mathbb F_1) = \ell(\mathbb F_2)=0$, if and only if both $\mathbb F_1$ and $\mathbb F_2$ are unramified at $p$. How would you express this as a relation between $\mathbb F_1$ and $\mathbb F_2$? $\endgroup$ Commented Oct 9, 2022 at 1:05
  • $\begingroup$ For instance, if $\ell(F)=a$, can we say something about the degree of $[\mathbb{Q}(\zeta_{p^a}):(F \cap \mathbb{Q}(\zeta_{p^a}))]$, where $\zeta_n$ is a primitive $n$-th root of unity? $\endgroup$ Commented Oct 9, 2022 at 1:12
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    $\begingroup$ That degree may be any divisor of $p^{a-1}(p-1)$. $\endgroup$ Commented Oct 9, 2022 at 1:21

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